NEET · Physics · STD 12 - 2. Electric potential and capacitance
A capacitor of \(2\,\, \mu F\) is charged as shown in the diagram. When the switch \(S\) is turned to position \(2,\) the percentage of its stored energy dissipated is ......\(\%\)

- A \(20\)
- B \(75\)
- C \(80\)
- D \(0\)
Answer & Solution
Correct Answer
(C) \(80\)
Step-by-step Solution
Detailed explanation
Initially, the energy stored in \(2 \,\mu F\) capacitor is \(U_{i}=\frac{1}{2} C V^{2}=\frac{1}{2}\left(2 \times 10^{-6}\right) V^{2}=V^{2} \times 10^{-6}\, \mathrm{J}\)
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